Let #y = f(x)# be a twice-differentiable function such that #f(1) = 2# and . What is the value of #(d^2y)/(dx^2)# at # x = 1#?

Question

William Abdalla · Accepted Answer

#(d^2y)/(dx^2) = 28# when #x=1#

Let #y = f(x)# be a twice-differentiable function such that #f(1) = 2# and . What is the value of #(d^2y)/(dx^2)# at# x = 1#? We have: #dy/dx = y^2+3# We know that #y=2# when #x=1# and so: #y=2 => dy/dx = 2^2+3=7# Differentiating the above equation (implicitly) wrt #x# we get: #(d^2y)/(dx^2) = 2ydy/dx # And so when #x=1# we have: #(d^2y)/(dx^2) = 2(2)(7) = 28#

Samantha Adkison · Answer

#28# After two derivations, the terms like #a x + b# disappear. Then if #(dy)/(dx)=y^2+3->(d^2y)/(dx^2)=2y y'=2y(y^2+3)# but #y(1)=2# so #(d^2y)/(dx^2)=2 xx 2(2^2+3)=28#

Luke Alvarez · Answer

undefined To find the value of $\frac{{d^2y}}{{dx^2}}$ at $x = 1$, we need to differentiate the function $f(x)$ twice with respect to $x$ and then evaluate it at $x = 1$.

Given $y = f(x)$, and $f(1) = 2$, we need to find $f''(1)$, the second derivative of $f(x)$ evaluated at $x = 1$.

We're provided with $f'(x) = 3x^2 - 6x + 2$.

Differentiating $f'(x)$ with respect to $x$ again, we get:
$$\frac{{d^2y}}{{dx^2}} = \frac{{d}}{{dx}}(3x^2 - 6x + 2) = 6x - 6.$$

Substitute $x = 1$ into $f''(x)$:
$$f''(1) = 6(1) - 6 = 6 - 6 = 0.$$

So, the value of $\frac{{d^2y}}{{dx^2}}$ at $x = 1$ is $0$.